Bradley j. nartowt; solid state 2 – final exam; Monday, April 22, 2013 Zero sound in Fermi liquid: The transport equation for the distribution function f k (r , t ) in a Fermi liquid 3
like He looks quite similar to the Boltzmann equation,
df k 1 [ f k ] f k(0 ) vk r fk k ( ) coll ( f ) coll ; f k ,k ~ ; 3 t d t f f k V TF k k k
(1.1)
Here, where k is the shift of the quasiparticle energy due to both external fields and the Landau interaction, and f (0) is the Fermi function, e.g., fk fk(0 ) f k . (a) Assume that the system is in the collisionless regime fo r frequencies of interest, i.e. neglect the right hand side of the equation. Fourier transform this equation with with respect to time and space to obtain ( q vk ) f k ( k F )q vk fkk f k 0; (1.2)
k
Here, fk f k (q, ) is the Fourier-transformed distribution-function, and f kk is the landau interaction. Fourier transform using the ansatz fk fk ei ( qx t ) , notice that k f k(0) ( k ) . Since f k is the probability that a particle in a system has po sition sition x and momentum k at time t , t , write the energy-shift in terms of the full non-equilibrium function. One thus Four ier transforms the dispersion’s dispersion’s departure from equilibrium, f E k k k k k fk k fkk f k fk f k f k f k
and assume ume the the dis dispersi rsion (1. (1.1), for ( f )coll 0 . This maps
d dt
(1.3)
i and r iq , and we have,
f k (0 ) i ( q x t ) i ( q x t ) i ( q x t ) 0 ( fk e ) v k r fk e f f e k k k t k k f k (0 ) i ( q x t ) i (q x t ) i fk e v k iq f k e f f e k k k k k f k (0 ) i ( q x t ) f f ie fk v k q fk k k kk k i ( q x t )
f k (0 ) 0 ( vk q) f k v q fkk f k k k k
(1.4)
(b) Assume that fkk F0 s / N 0 , i.e. the Landau interaction is isotropic and nonmagnetic. Show that the 1
restriction of f k to the Fermi sphere f k ˆ obeys, f kˆ
q v k
q v k
s 0
F
The landau symmetric order parameter is, 1
It is this condition which gives
f
d kˆ
4
d
F 0 s q vk
f 1
kˆ
4
f kˆ (q, k , );
(1.5)
fkk N 0 F F (cos ) s 0
s 0
(2 1) F
s
3 n
P (cos ); N 0
(1.6)
2 F
0
If the amplitude f k of the fluctuation is within k BT / F of the Fermi sphere, then that means t emperature is low. In turn, we write k f k(0) ( k ) , in which F at sufficiently-low temperature. Treating this ansatz (1.5) into (1.4), we have, 0 ( v k q)
d
F0 s q v k
f 1
kˆ
4
( k F ) vk q fkk k
F 0s q v k
f 1
ˆ k
d 4 (1.7)
d ( k F ) 1 s s (2 1) F P (cos ) F0 v k q fk ˆ 1 0 4 N 0 k q v 1 0 k
For (1.7) to be consistent, we need, ( k F )
N 0 / F 0 s
(1.8)
q v k k q vk
(c) Expand f k ˆ in Legendre polynomials on the Fermi sphere. f kˆ
q v k
q vk
s
F0 f kˆ
d
To use this self-consistency relation (1.9), notice that
4
Y
cos qvk
cos
s
F0 f k ˆ
d 4
(1.9)
Y d mm , so the integral of (1.9) is really an
* m m
inner-product over the basis-set of spherical functions; effecting orthonormality, we have, 1 F0 s f kˆ d 41 ( F0s P 0 )( fkˆ P ) d F0s fkˆ P0 (cos ) F0s fkˆ 1 F0s f 4 kˆ
(1.10)
Show there is a solution f k ˆ with isotropic symmetry only if,
1
2 qv F
1 F 0 s 1
ln(
qv F ) 0 qvF
(1.11)
Iterating the self-consistency relation (1.9) once, we have, f k ˆ
F0 s fkˆ F0s 4
F0 s fk ˆ F 0s
cos qvk
cos cos
cos qvk
cos qvk
cos
qv ln 2 k
qvk qvk
d
F0s fk ˆ F 0s 4
cos qvk
cos
1
2
x dx
1 qvk
x
1 1 1
(1.12)
This is not right. (d) Plot the solution ω as a function of qv F numerically, or sketch it at low qv F , and determine the behavior analytically. Using Mathematica’s ContourPlot command, we plot on the y-axis, and qv F on the x, for various values of F 0 s {0.2,1,3,5,8} , we get,
3.0
2.5
2.0
1.5
1.0
0.5
0.0
3
2
1
0
1
2
3
(1.13)
This is the dispersion for collisionless zero sound, a sound mode in a degenerate Fermi liquid which has no analog in the noninteracting system. There are two things to note: (1) the dispersion is actually linear, and (2) the velocity of the zero-sound is on the same order of magnitude as t he Fermi velocity.
Superconducting Sphere: Consider a type-I superconducting sphere with critical field Hc, placed in a uniform ˆ. external field H 1 B M H 0 z
(a) Show (with a simple argument) that the f ield at the surface of the sphere is parallel to the surface. The field at the surface of the sphere is parallel to the surface is due entirely to the Meissner effect. That is the “simple” answer to this question. We can formalize this reasoning. Semi-formalizing our reasoning: Consider a superconducting sphere of radius b a nd magnetic permeability (for generality), which allows the relation H 1 B to be well-defined2. Maxwell’s equations outside the sphere,
B 0 J 0 0 ( A) 2 A; B 0; The current-free space admits a magnetic scalar potential, with the following boundary conditions, uniform H-field M : H M ; M (b, ) 0; lim r b ( M ( r, )) M H 0 z H 0 r cos ;
(1.1)
(1.2)
It is the Meissner effect that gives us the boundary condition M (b , ) 0 , and the uniformity of the field that gives us lim r b ( M ( r, )) H 0 r cos . Hence, the field is parallel to the surface. We can even pro ve this. Formalizing our semi-formal reasoning: In plane-polar coordinates, Laplace’s equat ion has the solution that vanishes on a radius-b sphere,
M (r , ) A r B r
0
2
( 1)
A b 2 1 b2 1 P (cos ) A r r 1 P (cos ) A r r 1 P (cos ); (1.3) 0 0
In a perfect superconductor, 0 , as will be shown in Appendix II. However, consider both B and H to be finite.
We then further specialize (1.3) to the uniform-at-infinity boundary condition,
M (r b, ) H 0 r cos A r 0 P (cos ) H 0 r1P1 (cos ) A 1 0; A1 H 0
(1.4)
0
Thus, the magnetic scalar potential is M H 0 [1 (b / r )3 ]r cos , and the consequent field is, H M H 0 zˆ H 0 b ( 3
cos r2
) H 0 ( zˆ b
3
2rˆ cos sin ˆ r3
ˆ rˆ )rˆ m ˆ 3 3(m ) H 0 ( zˆ b ) r 3
(1.5)
ˆ we see that the field at the sphere-surface is solely due to the uniformlySetting r b , and dotting H(b, ) r applied field, H(b, ) H 0 ( zˆ 2rˆ cos sin ˆ) H 0 (rˆ cos sin ˆ) H(b, ) rˆ H 0 rˆ cos (1.6) (b) Use the previous fact to show that when the external field reaches a value of 2Hc/3, some part of the sample must become normal. At some field strength, the field at the surface H (b , ) must exceed the applied magnetic field. This must be when the d ipole-field (produced by a uniformly-magnetized sphere) begins to contribute to the applied field. Dotting (1.5) with the ˆ vector, and suppose H 0 H c caused the field to be as if b 0 ; we see, ˆ ˆ ˆ ˆ cos sin ˆ ˆ 3 2 r 3 0 sin ˆ ˆ H H 0 ( zˆ b ) H 0 ( sin b ) r3 r 3
r b 2 H sin b 0 H 0 sin (1 3 ) b 0 max H r b 2 H 0 r H 0 sin H C sin
(1.7)
3
Consider the inequality at the surface r = b, r b b3 1 1 1 1 ˆ 30 210 30 H H 0 H 0 sin (1 3 ) H 0 sin sin 3 r 1 (b / r ) 2
(1.8)
It must be along this angular interval that the superconducting state begins to be de stroyed at the surface. Subtracting the field at this angle from (1.7) for the case of max H r b 2 H 0 in which H 0 H C , we find the change in magnetic field at the surface before and after the superco nducting-to-nonsuperconducting transition, 1 3 2 H 0 max H r b (H ˆ)b0 ( r , 210) 2H C H C ( ) H C H C H 0 (1.9) 2 2 3 Computing the field inside and out side the sphere at H = (2/3)HC: Suppose the external applied field reached the ˆ at the surface. If our sphere had a uniform magnetic field H 23 H C z ˆ at r b , the value of H 23 H C z superconducting state would suffer a supercurrent in the reg ion r b that would generate a uniform field. The 3 field outside the sphere (due to the supercurrent) would be that of a dipole , where the “magnetization” would ˆ within r b , yielding a dipole moment of, be chosen to cancel 23 H C z H(r b)
3
2 3
H C M( r b)
2 3
H C zˆ
m V
8
m 4 3
b
3
m H C b3 zˆ
D. J. Griffiths, Introduction to Electrodynamics, p. 246, problem 5.33, Eq. [5.87], 3 rd ed.
9
(1.10)
4
Computing net the field outside the sphere , ˆ m 2 ˆ 0 3(m rˆ )r 0 3( 98 H C b3 zˆ rˆ) cos1 zˆ 98 H C b3 z ˆ H H ( r b) H 0 zˆ H z C 3 3 4 r 3 4 r
2 0 3( 98 b3 ) 98 b3 0 ( 34 b3 ) 2 ˆ ; z H C zˆ H C zˆ 3 3 3 4 r 3 2 r
(1.11)
nd
Now, compare (1.11) to the field outside a sphere with magnetic permeability , and note that the 2 term in (1.11) is precisely the field we get if we set 0 , H H uniform H dipole H 0 zˆ
0 3(m rˆ)rˆ m 4
r
3
4 b3 ( 0 ) H 0 2m ˆ (1.12) H0 z H ( r , ); m ; 3 4 r 2 0
(c) At what external field does the entire sphere become normal? Let the sphere be of radius b. Consider the Helmholtz free energies of the superco nducting vs. the nonsuperconducting states, 4 3 0 1 0 1 0 0 2 2 0 Fn Fn FB Vfn Vf B V fn H b fn H ; Fs F s ; 2 2 3
(1.13)
The critical field is computed by setting magnetic energ y in the sphere equal to the heat of transition, 4
1
3 F s
3
2
2 b 3
F s Fs0 Fn0 V f s0 fn0 b3 H C 2 H C
(1.14)
We compute the Helmholtz free energy o f the difference between t states using Ginzburg-Landau theo ry. Let the applied magnetic field be uniform and close to the critical field. Let t he temperature be close to the transition-temperature. Then, homogeneity 2 2 2 4 2 4 F s a 12 d 3 x aV 12 V ; 4m
2
Extremizing (1.15) with respect to (in which it is simply quadratic) yields F s 0
(1.15)
2
a / .
Identifying a (T T C ) , we have F s 2V 2 (T T C ) 2 , and consequently, H C
3 2
V
2
2 (T T C ) 2 3
b 0
(T T C )
V 4 b
V
3
0 34 b 3
3
(T T C ) 0
(1.16)
Appendix I – derivation of the London penetration depth to reinforce the argument that the field lines are tangential to the surface of a spherical superconductor
Preliminaries: consider (two of) Maxwell’s equations in the magnetostatic, charge-free space, B 0 J ( A) 2 A; B 0 ( B) 2B; B B B nˆ; This would appear as,
4
D. J. Griffiths, Introduction to Electrodynamics, p. 264, example 6.1, 3 rd ed.
(1.17)
(1.18) Let the type-I superconducting sphere have its quantum state (and its gauge-transform) appear as, q gauge-transform (r, t ) nei n exp i ( 0 ) ; 0 (rt ); c
(1.19)
5
Two currents: Superconducting current ( J s ) is defined to dissipate no Joule heat , and is associated with the superconducting state in equilibrium, whose velocity is v s , whose spatial density is n s , and consequent current, J J n J s ; J s qs ns v s
e 2
ns v s
e 2
ns
m
;
(1.20)
Gauge invariance: This current (1.20) needs to have its equations of motion (1.17) be invariant under gauge transform. To this end, the phase-gradient of (1.20) transforms as,
( ) ( 0 ) cq (rt ) cq (rt ) cq A;
(1.21)
Put (1.21) into (1.20), identify q 2e , regard the superconductor as perfectly non-magnetized, take J s 0 , and effect the curl upon the statement that results, e e e2 2e 2e 2 J s ns A ns B ns H J ( H) H (1.22) 0 2m 2m mc c c Approximation: Let us be within a distance o f a characteristic length , defined by (1.22) as 2 n s
e 2 mc
, and
consequently a field’s radial component should be exponentially attenuated within the superconductor. By the ˆ and nˆ boundary conditions B rˆ 0 H 0 zˆ rˆ 0 H 0 cos and H rˆ H 0 zˆ nˆ , we can assert that both the r components of the H-field should exponentially decay within
nS mc e
2
of the surface.
Appendix II – Macroscopic electrodynamics for field lines of a superconductor ***NOTE***: The following presentation is due to a problem statement given by Prof. Charles Thorn while teaching Electromagnetic Theory I, where field lines were plotted for a sphere in a uniform magnetic field.
A superconductor, macroscopically speaking, is perfectly no n-susceptible. We can see this show up in a plot of the field lines of a superconducting sphere. ˆ , Eq. (1.12). Breaking H into Consider the magnetic field outside a sphere o f radius b in a field H H 0 z components, we get H r cos (4 r 3 H 0 2m ) and H sin ( 4 r 3 H 0 m) ,
5
In contrast to the normal current ( jn ), associated solely with the evolution of Joule heat.
H( r , ) H 0 zˆ m
The field lines are given by
1 dr r d
ˆ 3 zr r 2 z 5
4 r
cos 2m 4 r3 H 0 rˆ sin m 4 r3 H 0 ;
(1.23)
E E , which uses the arclength formula and posits the triangle-congruence r
( Er , E ) ( dr, d ) . Consider us to be in the (r , ) 2D-plane of coordinates. which yields,
( br )3 2 4 Hm b 4 r 3 H 0 2m r cot cot r 3 3 ( ) 4 Hm b d (b ) b H b 4 r H m b 0 b dr
r H r
r
3
0
3
0
1 3 2 3
d cot d
d (sin ) sin
ln sin
1 3 2 3
3 2 cot 3
d ln
2
3
sin 2
(1.24) A0 2
3
Note that we introduced,
r b
;
m 4 H 0 b
3
;
(1.25)
6
Due to idiosyncrasies with Maple , we are only able to superimpose multiple plots if we convert to Cartesian coordinates. in the xz-plane, we have sin 2 x 2 x (careful! is “tilting outward from” the z-axis—it is x z
not a polar angle). Carefully massaging this by first isolating and then introducing x 2 z 2 , we get, 1 2/ 3
A0 1 A0 1 3 2 x ( x ) 1 z ( x ) 2 2 1 2 3 2 3 2 2 3/ 2 2 2 1 2 x ( x z ) x x
2
A0
A0
x 2 (1.26)
Let’s plot field lines not only for a perfectly non-suscept ible superconductor, but also for a perfectly susceptible sphere, for contrast. We need to figure out what
m 4 H 0b3
3
4 b ( 0 ) H 0 1 2 0 4 H 0b3
0 2 0
limiting cases: for 0 (perfectly-repelling of magnetic field lines) we have have
6
0 0
becomes in the two
01 02
21 . For , we
1,
When I was a lad in Electromagnetic Theory I, I hadn’t learned Mathematica yet, and was exclusively using Maple 9.5.
Chi = 1 – infinite “appetite” for field lines
chi = -1/2 – infinite repulsion of field lines
Anisotropic Heisenberg Ferromagnet: 1D Heisenberg-type model with nearest-neighbor only interactions, H
J S S z i
z
i
12 J ( Si S i1 S i S i1 )
z i 1
(1.1)
Basis-switch: effect a transform to canonical Dyson-Maleev boson-variables, [a j , a ] j ; these appear as,
S ( ai ai S ); Si 2 S ai (1 z i
ai ai 2 S
z† z † ); Si 2 S ai ; Si Si ; ( Si ) Si ;
(1.2)
The commutation-relations amidst7 the S , S z , despite appearances, are the same as the Primakoff-Holstein, [ Si , S j ] 2 ij S zi ; [ Siz , S j ] ij Si ; [ Siz , S j ] ij Si ;
(1.3)
Putting (1.2) into the Hamiltonian (1.1) and manipulating using the algebra (1.3), we treat S zi S iz 1 , Si S i1 , and Si S i1 separately, S zi Siz 1 ( ai ai S )( ai1 ai1 S ) 2 ( ai ai ai1 ai1 ai ai S Sai1 ai1 S2 );
Si Si1
Si Si 1
2S ai (1
2 S ai
ai ai 2 S
) 2 S ai 1 2 (2 Sai ai1 ai ai ai ai1 );
2 S ai1 (1
ai1ai 1 2 S
(1.4)
) (2 Sai ai1 ai ai1 ai1 ai 1 ); 2
Under (1.4), the Hamiltonian (1.1) appears as,
J a a a a a a S Sa a J [2a a a a ( a a a a
H 2
z
i
2
z
i
i
1 2
i
i 1 i1
i
i i 1 i 1
i
i1 i1
i
J J z
i
i
i 1
i
S 2 12 J 2 Sai ai1 ai ai ai ai1 2 Sai ai1 ai ai1 ai1 ai1 ai ai1 ai1 ai1 )]
S[ ( ai ai1 J J z
ai ai1 )
( ai ai
ai1 ai1 )]
S
2
(1.5)
Effect a spectrum-analysis of this Hamiltonian by writing a Fourier-decomposition, recalling we are in 1D, 1 1 i k x ik x j a j e j bk a j e bk ; [ bk , bk ] kk ; [ bk , bk ] 0 0 † [ bk† , bk† ]; x arˆ; (1.6) N k N k
Then, the various non-diagonal operators (transitions) in (1.5) appear as, 1 ai ai ai1 ai 1 2 ei ( k k k k )ia bk bk bk bk e i ( k k ) a ; N kk k k
ai ai 1ai 1ai1 ai1 ai ai ai
ai ai 1
ai ai1
1 2
N
i ( k k k k )ia
e
kk k k
ai ai 1
ai ai ai1 ai1
ai 1ai
1
e N kk
1
e
i ( k k k ) a
e
(1.7) i ( k k ) ia
N kk
i ( k k ) ia
e
ik a
i (k k )a
1 e
Putting (1.7) into (1.5), and summing over i to get Kronecker deltas 7
e ika bk bk bk bk ;
e
b b ; k
k
b b k
e i
ika
k
i ( )ia
, we have,
Barentzen and Wrobel of Max Planck Institute, Generalized Dyson-Maleev representation of doped and undoped Heisenberg ferromagnets, 1993.
1 i (2 k kk ) a J bk bk bk bk k k bk bk bk bk k k 2 cos ka 2 N 2e J z kk k 2 H J z SJ 2bk bk cos ka S ( nˆi nˆi 1 S ) J k i z 1 J i (2 k k k ) a bk bk bk bk k k cos ka 2 bk bk bk bk k k e N J kk k z 2 J z J 1 2 S k 2bk bk cos ka 2bk bk N i S NJ z 2
(1.8)
Throwing away non-quadratic terms: Let there be no quartic-interactions (1.8); since we are in k-space, it is not nearest neighbor interactions we are throwing away, but rather magnon/magnon interactions; we thus get, H 2
J z S
J 2 2 2 J S J S 2 b b cos ka 2 b b z k k z k k 2 Jz S i Jz k 1
For both the isotropic and anisotropic case,
J
S J 4 sin i
z
k
2
ka 2
bk bk
(1.9)
J S E [ground state energy] 0 . 2
i
z
0
(a) Show that for Jz > J ⊥ , the magnon dispersion E(q) is similar to the isotropic case, b ut with nonzero energy as k → 0. Introduce the dimensionless J / J z ; then, for 1 , we can plot the dispersion (1.9),
H i E 0 2
J z S
vs.
12 ka2 21 , and this appears as, Eka E 0 2 J Z S 3.0 0.2
0.5
2.5
0.8
2.0
1.5
1.0
0.5
ka 3
2
1
0
1
2
3
(1.10) We see that E 0 0 ; because this is positive and non-zero de finite, we have non-zero ground state energy. (b) Show for the opposite case Jz < J ⊥ the ferromagnetically ordered state with magnetization along z is not the ground state! Again, we can plot this for 4 , but this only yields a newly-scaled but completely-identical version of (1.10),
Eka E 0 2 J Z S 2
30
4
25 8
20
15
10
5
ka 3
2
1
0
1
2
3
(1.11) However, this was computed with respect to the ground-state energy,
J S E . We could take the 2
i
z
0
extreme limit, and realize the energy bands of (1.11) could be highly up-shifted, 2
Eka J Z S
30
25
20
15
10 2
4
5
8
ka 3
2
1
0
1
2
3
(1.12) I guess that means that ka → 0 doesn’t correspond to the ground state. Appendix I – computation of Dyson-Maleev algebra
It can eb shown that these satisfy the spin-commutation-relations. You will need,
[ AB, C] A[ B, C] [ A, C] B; [ A, BC] [ A, B] C B[ A, C]; [ a, a ] 1;
(1.13)
Then,
[S , S ] 2
2S
[a (1 1
a a 2S
), a] [ a , a]
1 2S
[ a a a, a] 1
a a [ a, a] [ a a , a] a 2S
1
0 (a [a , a] a [ a , a]) a 2S
(1.14)
[ S , S ] 2 2 S ( a a ) a 2 2 a a S 2 S z
2
z
[ S , S ]
2
[ a , a ]a a [ a , a ] [a , a a S ] [ a , a a] [ a a a, a a] 0 1 ([ a a a, a ]a a [ a a a, a]) 2S 2 S 2S 2 S 1 ( a a [ a, a ] [ a a , a ] a) a 1 a a [ a, a ] a a a (1.15) 2S a ( a a [ a, a] [ a a , a] a) 2 S a ( a a [ a, a ] [ a a , a]a) z
[S , S ]
1
2 S
a a a
a
1
1
1
a a a a (0 2 a a) a a a a 2S 2S
Then,
[S , S ] z
2
a a a a a 2 2 S a 2 S a 1 Si ; 2S 2 S
(1.16)
